Let $C\le A, D\le B$. Prove $C\times D=\{(c,d)\in A\times B:c\in C,d\in D\}$ is a subgroup of $A\times B$
I know that in order to prove this, that the following needs to be proved:
- $C\times D$ needs to be closed under it's group operation, if $a,b\in C\times D$ then $ab\in C\times D$
- There needs to be an identity element, $1\in C\times D$ such that $1a = a$ where $a\in C\times D$
- There needs to be an inverse element: $a\in C\times D$ then $a^{-1}\in C\times D$ and $aa^{-1} = 1$
I'm pretty sure that I need to do all of this in order to complete the question, but I'm not sure where to start and how to prove this.
Use the one-step subgroup test
Since $A\le C$ and $B\le D$, we have $e_C\in A$ and $e_D\in B$; hence $(e_C,e_D)\in A\times B$. Thus $A\times B\neq \varnothing$.
By definition, $A\times B\subseteq C\times D$.
Let $x,y\in A\times B$. Then there exist $a,s\in A$ and $b,t\in B$ such that $x=(a,b), y=(s,t)$. Now
$$\begin{align} xy^{-1}&=(a,b)(s,t)^{-1}\\ &=(a,b)(s^{-1},t^{-1})\\ &=(as^{-1}, bt^{-1}), \end{align}$$
which is in $A\times B$, since $as^{-1}\in A$ and $bt^{-1}\in B$ (as $A\le C$ and $B\le D$). So $xy^{-1}\in A\times B$.
Hence $A\times B\le C\times D$.